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Longitudinal Bending Stress
Stress at saddles
in ts 1 inA 40 Q 250000 lb
R 50 inK1 0.335
H 50 inL 1000 in
P 250 Psi
α
Ro
A = 40 in b = 24 in
L = 1000 in
Do = 100 in
H
α = 120o
E = 85 %
S = 17500 Psi (material SA-516, grade 65,
620 oF
Woperation = 500.000 lb
Q = lb
Stress at midspan
S1
Q A 11
AL
R2 H22 A L
14 H3 L
K1 R2 ts
S1 1.47 103 Psi
S1
Q L4
12 R2 H2
L2
14 H3 L
4 AL
3.14R2 ts
S1 7.619 103 Psi
Based upon our calculation, the stress at midspan is higher than the stress at saddles, it
results tension at the bottom and compression at the top.This is the following profile
resulting the higher midspan stress :
Stress due to internal pressure
Stress value of girth seam
Sum of tension
S1 + S2 = 15310 psi
S2P R2 ts
S2 7.692 103 Psi
Psi
For A = 40 in, we found that the sum of tension is higher than the stress value of girth
seam, it means that the vessel will collapse, or on the other words the value of A is still
not satisfactory.
We have to make some of trial and error to find the exact value of A.
Assume the value of A = 96 in
Stress at saddles
Stress at midspan
S1
Q A 11
AL
R2 H22 A L
14 H3 L
K1 R2 ts
S1 5.379 103 Psi
S1
Q L4
12 R2 H2
L2
14 H3 L
4 AL
3.14R2 ts
S1 5.424 103 Psi
Sum of tension
S1 + S2 = 13116 psi
The sum of tension doesn’t exceed the stress value of the girth seam. The value of A is
already satisfactory.
Tangential Shear Stress (S2)
Sice A (96) > R/2 (50/2), the applicable formula :
S2 doesn’t exceed the stress value of shell material multiplied by 0.8 ; 0.8* 15500
= 12400 psi
Circumferential Stress
S2K2 QR ts
L 2 A
L43
H
S2 5.459 103
Psi
Stress at the horn of saddle (S4)
Since L (1000) > 8.R (400), A (96) > R/2 (50/2)the applicable formula :
For the value of b = 24 in, S4 exceeds the stress value of shell material multiply by 1.5 ;
17500x1.5 = 26250 psi. the value of b is not satisfactory.
Assume that the value of b = 60 in. the value of S4 will be :
Since S4 doesn’t exceed the stress value of shell material multiply by 1.5, thus the value
of b = 60 in is already satisfactory.
Stress at bottom of shell (S5)
; where K7 = 0.76 ; yield point = 35000 psi (page :
147 ).
S4Q
4.ts b 1.56 R ts( )
3 K6 Q2 ts
S4 2.673 104
S4Q
4.ts b 1.56 R ts( )
3 K6 Q2 ts
S4 2.556 104
S5K7 Q
ts b 1.56 R ts( )
S5 3.343 103
Psi
psi
S5 doesn’t exceed the compression yield multiplied by 0.5 ; 35000x 0.5
= 17500 psi.
